Question

Помогите Пожалуйста!!!
1,2,3 номер​

Answer 1

#1

1)(х-8)*(х+4)

${x}^{2} - 4x - 32 = 0 \\ d = ( - 4) ^{2} - 4 \times 1 \times ( - 32) = 16 + 128 = 144 = {12}^{2} \\ x1 = \frac{4 + 12}{2} = \frac{16}{2} = 8 \\ x2 = \frac{4 - 12}{2} = \frac{ - 8}{2} = - 4$

2)(х-3)*(х-12)

$4 {x}^{2} - 15x + 9 = 0 \\ d = ( - 15)^{2} - 4 \times 4 \times 9 = 225 - 144 = 81 = {9}^{2} \\ x1 = \frac{15 - 9}{2} = \frac{6}{2} = 3 \\ x2 = \frac{15 + 9}{2} = \frac{24}{2} = 12$

#2

1)

${x}^{2} = y \\ y^{2} - 8y- 9 = 0 \\ d = ( - 8)^{2} - 4 \times 1 \times ( - 9) = 64 + 36 = 100 = {10}^{2} \\ y1 = \frac{8 - 10}{2} = \frac{ - 2}{2} = - 1 \\ y2 = \frac{8 + 10}{2} = \frac{18}{2} = 9 \\ x1 = - \\ x2 = \sqrt{9} = + - 3$

2)

$\frac{ {x}^{2} - 7x }{x + 2} = \frac{18}{x + 2} = \\ {x}^{2} - 7x = 18 \\ {x}^{2} - 7x - 18 = 0 \\ d = ( - 7)^{2} - 4 \times 1 \times ( - 18) = 49 + 72 = 121 = {11}^{2} \\ x1 = \frac{7 + 11}{2} = \frac{18}{2} = 9 \\ x2 = \frac{7 - 11}{2} = \frac{ - 4}{2} = - 2$

#3

$\frac{x - 1}{x + 2} + \frac{x + 1}{x - 2} + \frac{2x + 8}{4 - {x}^{2} } = 0 \\ \frac{x - 1}{x + 2} + \frac{x + 1}{x - 2} + \frac{2x + 8}{(2 - x) \times (2 + x)} = 0 \\ \frac{ {x}^{2} - x - 2x + 2 + {x}^{2} + x + 2x + 2 - 2x - 8}{(2 - x) \times (2 + x)} = 0 \\ \frac{2 {x}^{2} - 4 - 2x}{(2 - x) \times (2 + x)} = 0 \\ \frac{2 \times (x + 1) \times (x - 2)}{(x - 2) \times (x + 2)} = 0 \\ \frac{2 \times (x + 1)}{x + 2} = 0 \\ \frac{2x + 2}{x + 2} = 0 \\ 2x + 2 = 0 \\ 2x = - 2 \\ x = - 1$